Interaction by Jason Davies (drag the circles!)

Consider the olive triangle in (a). This is called a right angled triangle, because it has a square’s corner in it at the bottom right. The longest side is called the hypotenuse, and the two other sides are sometimes called the legs. There’s a long leg at the bottom and and a short leg to the right.

(a) (b) |

We want to prove that area of the square of the short leg and the square of the long leg is equal in combined area with the square of the hypotenuse. The red square in (b) is the square of the short leg, and the orange square (with the red square inside it) is the square of the long leg.

(c) (d) |

All of the coloured areas in (c) make up the square of the hypotenuse. To prove that together they’re the same area as the red and orange squares, the squares of the legs, think about how you can rearrange them, as in (d). You can see that they all fit together snugly in the square of the long leg, but to do so they have to overlap, and the size of the overlap is the square of the short leg.

Therefore the area of the square of the hypotenuse is equal to the area of the square of the long leg, plus an overlap the size of area of the square of the short leg. The statement that these diagrams prove is known as Pythagoras’ theorem.

© Jason Davies 2012.